# Proof

(related to Proposition: Definition of Integers)

Let $$(a,b),(c,d)$$ be ordered pairs of natural numbers. We will prove that the relation defined using addition of natural numbers. \[(a,b)\sim (c,d)\quad\Longleftrightarrow\quad\begin{cases}(a+h,b+h)=(c,d)& or\(a,b)=(c+h,d+h).\end{cases}]`

is an equivalence relation. In other words, the integers as equivalence classes $$[a,b]:=\{(c,d),~(c,d)\sim (a,b)\}$$ are well-defined.

### $$( i )$$ Reflexivity $$(a,b)\sim (a,b)$$

Clearly, $$(a+0,b+0)=(a,b)$$.

### $$( ii )$$ Symmetry $$(a,b)\sim (c,d)\Leftrightarrow(c,d)\sim(a,b)$$

If $$(a+h,b+h)=(c,d)$$, then $$(c,d)=(a+h,b+h)$$, otherwise if $$(a,b)=(c+h,d+h)$$, then $$(c+h,d+h)=(a,b)$$.

### $$( iii )$$ Transitivity: If $$(a,b)=(c,d)$$ and $$(c,d)\sim (e,f)$$, then $$(a,b)\sim (e,f)$$

For symmetry reasons $$( ii)$$, it is sufficient to consider only this case: By hypothesis, $$(a+h_1,b+h_2)=(c,d)$$ and $$(c+h_2,d+h_2)=(e,f)$$ for some $$h_1,h_2\in\mathbb N$$. Therefore $$(a+(h_1+h_2)),b+(h_1+h_2))=(e,f)$$.

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013